1. Formulas, Equations, and Moles
Chapter 3
Formulas, Equations, and Moles

2. 3 ways of representing the reaction of H2 with O2 to form H2O
A process in which one or more substances is changed into one or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what happens during a chemical reaction
3 ways of representing the reaction of H2 with O2 to form H2O
reactants
products
3.7

3. Conservation of Mass
Antoine Lavoisier (1743–1794): Showed that mass of products is exactly equal to the mass of reactants.

4. Balancing Chemical Equations 01
A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved.
reactants  products
limestone  quicklime + gas
Calcium carbonate  calcium oxide + carbon dioxide
CaCO3(s)  CaO(s) + CO2(g)

5. Balance the following equation
C2H6 + O2
CO2 + H2O

6. Balancing Chemical Equations
Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2CO2
NOT
C2O4
3.7

7. Balancing Chemical Equations
Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
1 carbon
on right
2 carbon
on left
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O
3.7

8. Balancing Chemical Equations
Balance those elements that appear in two or more reactants or products.
multiply O2 by 7 2
C2H6 + O2
2CO2 + 3H2O
2 oxygen
on left
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H O2
2CO2 + 3H2O 7 2
remove fraction
multiply both sides by 2
2C2H6 + 7O2
4CO2 + 6H2O
3.7

9. Balancing Chemical Equations
Check to make sure that you have the same number of each type of atom on both sides of the equation.
2C2H6 + 7O2
4CO2 + 6H2O
14 O (7 x 2)
14 O (4 x 2 + 6)
12 H (2 x 6)
12 H (6 x 2)
4 C (2 x 2)
4 C
Reactants
Products
4 C
12 H
14 O
3.7

10. Mass Relationships in Chemical Reactions

11. Atomic mass is the mass of an atom in atomic mass units (amu)
Micro World
atoms & molecules
Macro World
grams
Atomic mass is the mass of an atom in atomic mass units (amu)
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H = amu
16O = amu
1amu = X g
3.1

12. Atomic and Molecular Mass 02
The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes.
Mass of C = average of 12C (98.89%) and
13C (1.11%)
= x 12 amu x amu
= amu

13. Average atomic mass of lithium:
Natural lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
Average atomic mass of lithium:
7.42 x x 7.016
100
= 6.94 amu
3.1

14. Average atomic mass (6.941)

15. Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that
contains as many elementary entities as there
are atoms in exactly grams of 12C
1 mol = NA = x 1023
Avogadro’s number (NA)
3.2

16. atomic mass (amu) = molar mass (grams)
eggs
shoes
Molar mass is the mass of 1 mole of in grams
marbles
atoms
1 mole 12C atoms = x 1023 atoms = g
1 12C atom = amu
1 mole 12C atoms = g 12C
1 mole lithium atoms = g of Li
For any element
atomic mass (amu) = molar mass (grams)
3.2

17. 1 amu = ? g 1 12C atom 12.00 amu 12.00 g 6.022 x 1023 12C atoms =
1.66 x g
1 amu x
1 amu = 1.66 x g or 1 g = x 1023 amu
x g M
= molar mass in g/mol
NA = Avogadro’s number
3.2

18. Do You Understand Molar Mass?
How many atoms are in g of potassium (K) ?
1 mol K = g K
1 mol K = x 1023 atoms K
1 mol K
39.10 g K x x
6.022 x 1023 atoms K
1 mol K =
0.551 g K
8.49 x 1021 atoms K
3.2

19. Atomic and Molecular Mass
The mass of a molecule is just the sum of the masses of the atoms making up the molecule.
m(C2H4O2) = 2·mC + 4·mH + 2·mO
= 2·(12.01 amu) + 4·(1.01 amu) + 2·(16.00 amu)
= amu

20. molecular mass (amu) = molar mass (grams)
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2 1S
32.07 amu 2O
+ 2 x amu
SO2
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = amu
1 mole SO2 = g SO2
3.3

21. Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O
1 mol C3H8O molecules = 8 mol H atoms
1 mol H = x 1023 atoms H
1 mol C3H8O
60 g C3H8O x
8 mol H atoms
1 mol C3H8O x
6.022 x 1023 H atoms
1 mol H atoms x =
72.5 g C3H8O
5.82 x 1024 atoms H
3.3

22. Molar Mass
Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

23. Stoichiometry 01
Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.

24. Stoichiometry
Yields of Chemical Reactions: If the actual amount of product formed in a reaction is less than the theoretical amount, we can calculate a percentage yield:

25. How Many Cookies Can I Make?
Limiting Reactant
How Many Cookies Can I Make?
You can make cookies until you run out of one of the ingredients
Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

26. How Many Cookies Can I Make?
In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

27. Limiting Reagents
6 red left over
6 green used up
3.9

28. Stoichiometry
In this figure, what is the limiting reagent?

29. Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al
mol Al
mol Fe2O3 needed
g Fe2O3 needed OR
g Fe2O3
mol Fe2O3
mol Al needed
g Al needed
1 mol Al
27.0 g Al x
1 mol Fe2O3
2 mol Al x
160. g Fe2O3
1 mol Fe2O3 x =
124 g Al
367 g Fe2O3
Start with 124 g Al
need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent
3.9

30. Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al
mol Al
mol Al2O3
g Al2O3
2Al + Fe2O Al2O3 + 2Fe
1 mol Al
27.0 g Al x
1 mol Al2O3
2 mol Al x
102. g Al2O3
1 mol Al2O3 x =
124 g Al
234 g Al2O3
3.9

31. Solution Concentrations 01
Concentration allow us to measure out a specific number of moles of a compound by measuring the mass or volume of a solution.

32. Solution Stoichiometry
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
M KI
M KI
volume KI
moles KI
grams KI
1 L
1000 mL x
2.80 mol KI
1 L soln x
166 g KI
1 mol KI x
500. mL
= 232 g KI
4.5

33. 4.5

34. Solution Concentrations 04
Dilution:
Is the process of reducing a solution’s concentration by adding more solvent.

35. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
after dilution (f) =
MiVi
MfVf =
4.5

36. How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00
Mf = 0.200
Vf = 0.06 L
Vi = ? L
Vi =
MfVf Mi =
0.200 x 0.06
4.00
= L = 3 mL
57 mL of water
3 mL of acid
= 60 mL of solution +

37. Solution Stoichiometry uses molarity as a conversion factor between volume and moles of a substance in a solution.

38. Equivalence point – the point at which the reaction is complete
Titrations
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to acid with unknown
concentration
UNTIL
the indicator
changes color
4.7

39. What volume of a 1.420 M NaOH solution is
Required to titrate mL of a 4.50 M H2SO4
solution?
WRITE THE CHEMICAL EQUATION!
H2SO4 + 2NaOH H2O + Na2SO4 M
acid rx
coef. M
base
volume acid
moles acid
moles base
volume base
4.50 mol H2SO4
1000 mL soln x
2 mol NaOH
1 mol H2SO4 x
1000 ml soln
1.420 mol NaOH x
25.00 mL
= 158 mL

40. Percentage Composition 01
Percent Composition: Identifies the elements present in a compound as a mass percent of the total compound mass.
The mass percent is obtained by dividing the mass of each element by the total mass of a compound and converting to percentage.

41. Percent composition of an element in a compound
C2H6O
%C =
2 x (12.01 g)
46.07 g
x 100% = 52.14%
%H =
6 x (1.008 g)
46.07 g
x 100% = 13.13%
%O =
1 x (16.00 g)
46.07 g
x 100% = 34.73%
52.14% % % = 100.0%

42. Empirical Formula 01
The empirical formula gives the ratio of the number of atoms of each element in a compound.
Compound Formula Empirical Formula
Hydrogen peroxide H2O2 OH
Benzene C6H6 CH
Ethylene C2H4 CH2
Propane C3H8 C3H8

43. Empirical Formula 02
A compound’s empirical formula can be determined from its percent composition.
A compound’s molecular formula is determined from the molar mass and empirical formula.

44. Empirical Formula 03
Combustion analysis is one of the most common methods for determining empirical formulas.
A weighed compound is burned in oxygen and its products analyzed by a gas chromatogram.
It is particularly useful for analysis of hydrocarbons.

45. g of O = g of sample – (g of C + g of H) 4.0 g O = 0.25 mol O
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
g CO2
mol CO2
mol C
g C
6.0 g C = 0.5 mol C
g H2O
mol H2O
mol H
g H
1.5 g H = 1.5 mol H
g of O = g of sample – (g of C + g of H)
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O

46. How to “Read” Chemical Equations
2 Mg + O MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
3.7

47. Methanol burns in air according to the equation
2CH3OH + 3O CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
1 mol CH3OH
32.0 g CH3OH x
4 mol H2O
2 mol CH3OH x
18.0 g H2O
1 mol H2O x =
209 g CH3OH
235 g H2O
3.8